Physics A
Unit: G482: Electrons, Waves and Photons
1(a) Name the charge carriers responsible for electric current in a metal and in an
electrolyte.
[2]
Candidate style answer
Examiners commentary
Electrons are the charged particles in a
metal and protons in an electrolyte .
Part(a) only scores one mark as the candidate
has muddled protons and ions.
(b)(i) Define electrical resistivity.
[2]
Candidate style answer
Examiners commentary
The electrical resistivity of a substance is
the resistance of a unit cube of it....
Part(b)(i) scores zero as this is not the
definition.
(ii) Explain why the resistivity rather than the resistance of a material is given in tables
of properties of materials.
[1]
Candidate style answer
Examiners commentary
Each resistor may have a different shape
and size giving it a different value but
the resistivity is always the same
part (ii) gains the mark with a b.o.d. (benefit of
the doubt) as the key idea of dimensions is
included.
Fig. 1.1
(c) Fig. 1.1. shows a copper rod of length l = 0.080m, having a cross-sectional area
A = 3.0 x 10-4 m2.
The resistivity of copper is 1.7 x 10-8 Ω m.
(i)
Calculate the resistance between the ends of the copper rod.
[2]
Candidate style answer
Examiners commentary
R = ρl/A
= 1.7 x 10 x 0.08/3.0 x 10
-8
-4
= 4.53 x 10-6
Resistance = …..4.5 x 10-6... Ω
OCR Level X in Physics A
Part(c) (i) shows that the candidate is aware of
the defining equation but no retrospective mark
for (b)(i) should be considered. One mark is
for correct substitution of the figures and one
for the correct answer.
1
(ii) The copper rod is used to transmit large currents. A charge of 650 C passes along
the rod every 5.0 s. Calculate
1
the current in the rod
[2]
Candidate style answer
Examiners commentary
I = Q/t
Part(c)(ii) is a U grade question where just the
answer with no working will score both marks.
There is no ambiguity so the candidate can
only have arrived at the answer by the correct
calculation. However it is recommended that
the candidate shows working to gain credit if
the calculation is performed incorrectly.
= 650/5
current = ………130..............A
2
the total number of electrons passing any point in the rod per second.
[2]
[Total:11]
Candidate style answer
Examiners commentary
number = ........ 8.1 x 1020........
Comments: The candidate gained 8/11 marks for this first question. The question is typical of
an opening question to help candidates settle into the examination. It relies on recall of basic
facts and definitions and simple calculations.
2(a) (i) Use energy considerations to distinguish between potential difference (p.d.) and
electromotive force (e.m.f.).
[2]
Candidate style answer
Examiners commentary
The potential difference across a resistor
is the energy heating it when 1 coulomb
of charge passes. The e.m.f. is the energy
given to each coulomb of charge moving
round the circuit by the battery…
In part (a) (i) the examiner is looking for per
unit charge or equivalent. As written this
candidate could mean V = EQ or V = E/Q. As
the same ambiguity occurs twice the penalty is
just one mark.
(ii)
Here is a list of possible units for e.m.f. or p.d.
J s-1
State which one is a correct unit:
J A-1
J C-1
[1]
Candidate style answer
Examiners commentary
J C-1
Part (a) (ii) is correct.
2
(b) Kirchhoff’s second law is based on the conservation of a quantity. State the law and
the quantity that is conserved.
[2]
Candidate style answer
Examiners commentary
The e.m.f. of the battery is equal to the
sum of potential differences across all of
the components in the circuit.
In part (b) the statement of Kirchhoff’s second
law is considered to be good enough but the
wrong quantity is conserved. Charge is the
conserved quantity for the first law.
Charge is conserved.
(c) A battery is being tested. Fig. 2.1 shows the battery connected to a variable resistor
R and two meters.
Fig. 2.1
The graph of Fig.2.2 shows the variation of the p.d. V across the battery with the current I
as R is varied.
(i)
Draw the line of best fit on Fig. 2.2.
[1]
Candidate style answer
Examiners commentary
In part (c) the straight line is adequate and the
intercept with the y-axis could be 6.1.
However there is no initial discussion or
algebraic formula given for the internal
resistance so one mark is lost.
Fig. 2.2
3
(ii)
Use your line of best fit to determine
the e.m.f. ε of the battery
the internal resistance r of the battery. Show your working clearly.
[3]
Candidate style answer
Examiners commentary
ε = ……6.1........ V
r. = (6.1 – 0)/2.0
r = …..3.05 .......Ω
(d) The variable resistor R is adjusted to give the values at point M on Fig. 2.2.
Calculate
(i)
the resistance of R at this point
[3]
Candidate style answer
Examiners commentary
At point M, I = 0.6 A and V = 4.3 V
In part (d) the candidate takes the value of the
p.d. on the line rather than at M so loses one
mark but the remainder of the question is
correct. With error carried forward applied the
candidate just loses one mark.
R = V/I = 4.3/0.6
R = …..7.2......Ω
(ii)
the power dissipated in R.
[2]
[Total: 15]
Candidate style answer
Examiners commentary
P = IV= 0.6 x 4.3 = 2.58
power =..... 2.6… W
Comments: The candidate scores 11/15 for this question.
4
3
Fig. 3.1 shows how the resistance of a thermistor varies with temperature.
Fig. 3.1
(a) (i) Describe qualitatively how the resistance of the thermistor changes as the
temperature rises.
[1]
Candidate style answer
Examiners commentary
the resistance decreases.
(ii)
The change in resistance between 80 oC and 90 oC is about 15 Ω.
State the change in resistance between 30 oC and 40 oC.
[1]
Candidate style answer
Examiners commentary
105 Ω...
(iii) Describe, giving a reason, how the sensitivity of temperature measurement using
this circuit changes over the range of temperatures shown on Fig. 3.1.
[2]
Candidate style answer
Examiners commentary
Near room temperature the change in
resistance is large but at much higher
temperatures it is small...
Part(a)(iii) was just enough to gain a mark
although there was no reference to equal
increments of temperature for each change in
resistance.
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(b) Fig 3.2 shows a temperature sensing potential divider circuit where this thermistor
may be connected, between terminals A and B, in series with a resistor.
(i)
Draw the circuit symbol for a thermistor on Fig. 3.2 in the space between terminals A
and B.
[1]
(ii) A voltmeter is to be connected to the circuit to indicate an increasing p.d. when the
thermistor detects an increasing temperature. On Fig. 3.2, draw the circuit connections
for a voltmeter to measure a p.d. that rises with increasing temperature.
[1]
Candidate style answer
Examiners commentary
In part (b)(ii) the voltmeter was placed
incorrectly and in
Fig. 3.2
(iii) The value of the resistor in Fig. 3.2 is 200 Ω. The thermistor is at 65 oC. Use data
from Fig. 3.1 to show that the current in the circuit is about 0.02 A.
[3]
Candidate style answer
Examiners commentary
the thermistor resistance is 105 Ω so the
total resistance is R = 305 Ω
I = V/R = 6/305 = 0.02 A
(iii) the candidate loses one mark for not
showing how 0.02 was achieved. The mark
would have been given for 0.0197 or 0.020
indicating that the calculation was actually
done.
(iv)
Calculate the p.d. across the 200 Ω resistor at 65 oC.
[1]
Candidate style answer
V = 0.02 x 200
p.d. across resistor = ……4.0........V
6
Examiners commentary
(c) The graphs X, Y and Z in Fig 3.3. show how the p.d. across the resistor varies with
temperature, for three different values of the resistor.
Fig. 3.3
(i)
The values of resistance used are 20 Ω, 200 Ω and 1000 Ω.
State, explaining your reasoning clearly, which graph, X, Y or Z, is the curve for the
1000 Ω resistor.
[2]
Candidate style answer
Examiners commentary
At 20 oC the resistance of the thermistor
is 500 Ω. As the resistor is 1000 Ω a ratio
of 4 V to 2 V is needed with the bigger
voltage across the resistor so X is the
graph.
(ii) State one advantage and one disadvantage of using output Z for the temperature
sensing circuit.
[2]
[Total: 14]
Candidate style answer
Examiners commentary
advantage .the change in resistance and
the change in resistance are almost
proportional
In part (c)(ii) the first marking point was given
with a b.o.d. although the line is not drawn to
the origin. The second point made was
equally if not more vague; the range of p.d.s is
too small would have been adequate. The
second mark was not given.
disadvantage the values of potential
difference are too small
Comments: The candidate scored 10/14 for this question.
7
4(a) Fig.4.1 shows the electromagnetic spectrum.
visible
A
X-rays
u.v.
B
microwaves
radiowaves
--------------------------------------------------------→ increasing wavelength
In the spaces in Fig. 4.2, identify the principal radiations A and B and for each suggest a
typical value for the wavelength.
[4]
Candidate style answer
principal radiation
A
B
Examiners commentary
λ/m
gamma-rays
10-9
infra-red
10-6
In part (a) the wavelength of gamma rays is
outside the range losing one mark.
Fig. 4.2
(b)
State two features common to all types of radiation in the electromagnetic spectrum.
[2]
Candidate style answer
Examiners commentary
They can travel through a vacuum and
they can be polarised.
(c)
(i) Define the term plane-polarisation of visible light waves.
[1]
Candidate style answer
Examiners commentary
The light wave can only exist in one
plane..
In part (c)(i) the answer is too vague or
ambiguous to earn the mark
(ii)
Explain why sound waves cannot be plane-polarised.
[2]
Candidate style answer
Examiners commentary
Sound waves are longitudinal waves so
cannot be polarised.
(ii) lacks sufficient detail to be worthy of more
than 1 mark.
8
(d) Fig. 4.3 shows a student observing a parallel beam of plane-polarised light that has
passed through a polarising filter.
Fig. 4.3
(i)
Fig. 4.4. shows how the intensity of the light reaching the student varies as the
polarising filter is rotated through 360o in its own plane.
Fig. 4.4
Suggest why there is a series of maxima and minima in the intensity.
[2]
Candidate style answer
Examiners commentary
The polarising filter cuts out the light
completely at one angle of rotation but
lets it through at 90o to this angle.
The same comment is true for part (d)(i).
(ii)
Hence explain how sunglasses using polarising filters reduce glare.
[2]
Candidate style answer
Examiners commentary
(e) State an example of plane-polarisation that does not involve visible light and state
how the polarised wave may be detected.
[2]
[Total: 15]
Candidate style answer
Examiners commentary
Microwaves can be polarised. Dish
aerials can be used to detect microwaves
Part (e) fails to score any marks because there
is no context to relate the wave to polarisation.
Some statement about the rotation of the
transmitter and or receiver of the 3 cm wave
9
apparatus commonly available in the school
laboratory would have scored both marks.
Comments: The candidate showed that his knowledge of this subject is limited. The answers
lack detail. In fact there is no attempt at part (d)(ii). The score is 7/15.
5(a) State and explain one difference between a progressive and a standing wave.
[2]
Candidate style answer
Examiners commentary
The amplitude of a progressive wave is
the same along the wave but a standing
wave has places where the amplitude is
always zero called nodes
Part (a) gains both marks as a valid quantity
chosen and a comparison given.
(b) In an investigation of standing waves, a loudspeaker is positioned above a long pipe
containing water, causing sound waves to be sent down the pipe. The waves are reflected
by the water surface. The water level is lowered until a standing wave is set up in the air
in the pipe as shown in Fig. 5.1. A loud note is heard. The water level is then lowered
further until a loud sound is again obtained from the air in the pipe. See Fig. 5.2.
Fig 5.1
Fig. 5.2
The air at the open end of the pipe is free to move and this means that the antinode of the
standing wave is actually a small distance c beyond the open end. This distance is called
the end correction.
A student writes down the following equations relating the two situations shown.
l1 + c = λ/4
l2 + c = 3λ/4
(i)
Draw the standing wave in the pipe shown in Fig. 5.2 which corresponds to the
equation l2 + c = 3λ/4.
(ii)
10
[1]
On your diagram, label the positions of any displacement nodes and antinodes with
the letters N and A respectively.
[1]
Candidate style answer
Examiners commentary
In part (b) the sketch is correct but the bottom
N is forgotten at the water level losing one
mark.
(iii)
Use the two equations to show that l2 - l1 = λ/2.
[1]
Candidate style answer
Examiners commentary
l2 + c – (l1 + c) = 3 /4 - λ/4
giving l2 - l1 = λ/2
(iv)
The following results were obtained in the experiment.
frequency of sound = 500Hz l1 = 0.170 m l2 = 0.506 m
Calculate the speed of sound in the pipe.
[3]
Candidate style answer
Examiners commentary
λ/2 = 0.506 – 0.170
In (iv) there is no working. Either the figure
500 x 0.67 or the fuller answer of 335 must be
shown for the third mark.
λ = 0.67 m
speed =…..340……..m s-1
(c) The student repeats the experiment, but sets the frequency of the sound from the
speaker at 5000 Hz.
Suggest and explain why these results are likely to give a far less accurate value for the
speed of sound than those obtained in the first experiment.
In your answer, you should make clear the sequence of steps in your argument.
[4]
[Total: 12]
Candidate style answer
Examiners commentary
11
At a higher frequency the wavelength of
the standing wave in the tube will be
smaller so it will be more difficult to
measure and the calculation of the
speed will be less accurate.
In part (c) it is the measurement that is less
accurate rather than being more difficult so the
candidate has missed the point – given 2/4 for
recognising smaller wavelength and smaller
distances to measure.
Comments: The candidate scores 8/12.
6(a) Explain what is meant by the principle of superposition of two waves.
[2]
Candidate style answer
Examiners commentary
When two waves meet they add together
to give a new wave with a bigger
amplitude
In part (a) the mark is for the waves meeting.
(b) In an experiment to try to produce an observable interference pattern, two
monochromatic light sources, S1 and S2, are placed in front of a screen, as shown in
Fig. 6.1.
Fig. 6.1.
(i)
In order to produce a clear interference pattern on the screen, the light sources
must be coherent. State what is meant by coherent.
[2]
Candidate style answer
Examiners commentary
They are always in phase
In part (b)(i) the candidate focuses on phase
but not in sufficient detail so again scores one
mark.
(ii) In Fig 6.1, the central point O is a point of maximum intensity. Point P is the position
of minimum intensity nearest to O. State, in terms of the wavelength λ, the magnitude of
the path difference S1P and S2P.
[1]
Candidate style answer
λ/2.
12
Examiners commentary
(c) In another experiment, a beam of laser light of wavelength 6.4 x 10-7 m is incident on
a double slit which acts as the two sources in Fig. 6.1.
(i)
Calculate the slit separation a, given that the distance D to the screen is 1.5 m and
the distance between P and O is 4.0 mm.
[3]
Candidate style answer
Examiners commentary
using λ = ax/D
In part (c) the candidate thinks that the fringe
width x in the formula is from maximum to
minimum so loses one mark in (i) and despite
the possibility of error carried forward
a = λD/x
so a = 6.4 x 10-7 x 1.5/4.0 x 10-3
a = ……2.4 x 10-4.m
(ii) Sketch on the axes of Fig. 6.2 the variation of the intensity of the light on the screen
with distance y from O.
[1]
[Total:10]
Candidate style answer
Examiners commentary
gains no marks in (ii) because (b)(ii) and (c)(i)
give the information that the first minimum is at
4.0 mm
Fig. 6.2
Comments: This is an example of a question which should take the candidate a shorter time to
answer because there is a sketch, a simple calculation to complete and relatively little writing to
do.
The candidate scores 5/10.
7(a) The concept of the photon was important in the development of physics throughout
the last century. Explain what is meant by a photon.
[1]
Candidate style answer
A photon is a packet of light energy..
Examiners commentary
All three marks awarded for parts(a) and (b)(i);
(b) A laser emits a short pulse of ultraviolet radiation. The energy of each photon in the
beam is 5.60 x 10-19 J.
(i)
Calculate the frequency of an ultraviolet photon of the laser light.
[2]
Candidate style answer
Examiners commentary
E = hf so f = 5.60 x 10-19 /6.63 x 10-34
All three marks awarded for parts(a) and (b)(i);
frequency = ...... 8.45 x 1014 Hz
13
(ii) A photon of the laser light strikes the clean surface of a sheet of metal. This causes
an electron to be emitted from the metal surface.
1
The work function energy of the metal is 4.80 x 10-19 J. Define the term work
function energy.
[1]
Candidate style answer
2
Examiners commentary
no knowledge of the p.e. effect shown in (b)(ii).
Show that the maximum kinetic energy of the emitted electron is 8.0 x10-20 J.
[1]
Candidate style answer
(iii)
Examiners commentary
Show that the maximum speed of emission of an electron is about 4 x 105 m s-1.
[2]
Candidate style answer
Examiners commentary
k.e = ½mv
2
8.0 x 10
-20
= ½(9.1 x 10
v = 16 x 10 / 9.1 x 10
v = 4.2 x 105 m s-1
2
(c)
-20
-31
-31
)v
2
Part (b) (iii) has been completed by using the
figures given in the stem of the question and
this time working has been shown to score
both marks.
(i) State the de Broglie equation. Define any symbols used.
[2]
Candidate style answer
Examiners commentary
8 = h/mv.
The candidate has chosen the correct equation
from the data sheet so scores one easy mark
in part (c)(i) but fails to identify the symbols or
go on to do any calculation.
(ii) Calculate the minimum de Broglie wavelength associated with an electron emitted in
(b) above.
[2]
[Total: 11]
Candidate style answer
Examiners commentary
Comments: The candidate has scored 6/11 on this question. Either he/she decided to attempt
the last question first of short of time or else this is an area of the specification where knowledge
is weak.
8
The concept of energy is important in many branches of physics. Energy is usually
measured in joules, but sometimes the kilowatt-hour (kW h) and the electron volt (eV) are
more convenient units of energy.
Define the kilowatt-hour and the electron volt and determine their values in joules.
14
Suggest why the kilowatt-hour and electron volt may be more convenient than joules.
In your answer you should make clear how your suggestions link with the evidence.
Illustrate your answer by determining the energy dissipated by a 100 W filament lamp left
on for 12 hours and the kinetic energy of an electron accelerated through a p.d. of 1.0 MV
in a particle accelerator.
[12]
Paper Total [100]
Candidate style answer
Examiners commentary
1 kW h is a unit of energy. It is equal to
3.6 x 106 J.
1 eV is the energy given to an electron
when it is accelerated through a
potential difference of 1 V. It is equal to
1.6 x 10-19 J.
The kilowatt-hour is the unit used on
home electricity bills.
If a 100 W lamp is left on for 12 hours
then it uses 1.2 kW h of electricity using
the formula E = Pt. This is 4.3 x 106 J of
electrical energy.
The candidate scores a total of 6/12 for this
question. It appears that he/she has not
stopped to read the question properly and has
just picked a few sentences on which to base
an answer. The answer is also very short
possibly indicating a lack of time. The quality
of written communication mark has not been
awarded although what has been written is to
the point and is correct.
The first line gains 1 mark for the value of 1
kW h; The second and third lines 2 marks. The
use of the kW h gains 1 mark and the final
paragraph 2 marks.
The candidate has scored a mark 60/100 on
the unit.
The candidate has thrown away a number of
marks and with a little more forethought could
have achieved a better grade. The paper
shows promise and with further experience
and application this candidate could progress
to a high grade at A2 level.
15